Generalized Fermat's Little Theorem

Fermat's Little Theorem says that if p is prime and if p and a are coprime then p divides (a^p-a). Now if n is a positive integer is there a function of n f(n) such that if a and n are coprime then p divides (a^f(n)-a)? I have worked on some examples and from the fact that f(p)=p-1 p a prime I think that f(n) is the number of positive integers less than n and coprime to it, but how can it be proved? Note the number of positive integers less than n and coprime to it for the kth power of a prime are (p^(k-1))(p-1).

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Note by Samuel Queen
8 years, 4 months ago

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Comments

Your hunch is correct provided a typo in your statement is corrected. The statement should read n divides af(n)−1n \text{ divides }a^{f(n)}-1 or f(n)=1+Number of numbers relatively prime to n and less than n.f(n) = 1 + \text{Number of numbers relatively prime to n and less than n}. The fact that, if f(n)f(n) is the number of relatively prime numbers less than nn and if gcd⁡(a,n)=1\gcd(a,n)=1, then nn divides af(n)−1a^{f(n)}-1, is known as Euler's theorem. Calvin has a wonderful blog post on this. The function f(n)f(n) is called the Euler's totient function and is usually denoted as ϕ(n)\phi(n).

Marvis Narasakibma - 8 years, 4 months ago

Actually Euler's totient function serves as a generalisation to Fermat's Little theorem as this theorem is often referred to as.Since totient function for any prime number p equals (p-1), Fermat's theorem gives us a special case of Euler's theorem

Jaydutt Kulkarni - 8 years, 3 months ago
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